JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Quadratic Equations and Inequalities

Factor the radicands: $\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}=\sqrt{4(x-3)\left(x-\frac{1}{2}\right)}$, which is defined only for $x\le -3$ or $x\ge 3$. At $x=3$ every term is zero, so $x=3$ is a root. For $x>3$, dividing through by $\sqrt{x-3}$ leaves $\sqrt{x-1}+\sqrt{x+3}=2\sqrt{x-\frac{1}{2}}$; squaring twice reduces this to $6x=7$, i.e. $x=\frac{7}{6}$, which lies outside the domain and is rejected. For $x\le -3$ the right side stays strictly greater than the left, so no root occurs there. The equation therefore has exactly one real root.